A plane electromagnetic wave of frequency $25 ext{ MHz}$ travels in free space along the x-direction. At a particular point in space and time, the electric field is $\vec{E} = 6.3 \hat{j} ext{ V/m}$. (a) Calculate the magnitude and direction of the magnetic field vector $\vec{B}$ at this point. (b) Calculate the energy density of the electromagnetic wave.

Q: A plane electromagnetic wave of frequency $25 \text{ MHz}$ travels in free space along the x-direction. At a particular point in space and time, the electric field is $\vec{E} = 6.3 \hat{j} \text{ V/m}$. (a) Calculate the magnitude and direction of the magnetic field vector $\vec{B}$ at this point. (b) Calculate the energy density of the electromagnetic wave.

Given Data: Frequency of the electromagnetic wave ($\nu$) = $25 \text{ MHz} = 25 \times 10^6 \text{ Hz}$ Direction of propagation = x-direction ($i$) Electric field vector ($\vec{E}$) = $6.3 \hat{j} \text{ V/m}$ Permeability of free space ($\mu0$) = $4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$ Permittivity of free space ($\varepsilon0$) = $8.854 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)$ Speed of light ($c$) = $3 \times 10^8 \text{ m/s}$ Part (a): Magnitude and Direction of Magnetic Field Vector ($\vec{B}$) Magnitude Calculation: The relationship between the amplitude of the electric field ($E0$) and magnetic field ($B0$) is given by: $$B0 = \frac{E0}{c}$| Substituting the given values: $$B0 = \frac{6.3 \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 2.1 \times 10^{-8} \text{ Tesla (T)}$$ Direction Calculation: The direction of propagation of an electromagnetic wave is given by the direction of the cross product vector $\vec{E} \times \vec{B}$. Since the wave travels along the x-axis ($+\hat{i}$) and the electric field is along the y-axis ($+\hat{j}$): $$\hat{i} = \hat{j} \times \hat{k}$| Therefore, the magnetic field $\vec{B}$ must point along the z-axis ($+\hat{k}$). Thus, $\vec{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}$. Part (b): Energy Density of the Electromagnetic Wave\nThe total energy density ($u$) of an electromagnetic wave is the sum of electric energy density ($uE$) and magnetic energy density ($uB$), which are equal to each other ($uE = uB$). Formula for Energy Density: $$u = \varepsilon0 E0^2$$ Calculation: $$u = (8.854 \times 10^{-12}) \times (6.3)^2$$ $$u = 8.854 \times 10^{-12} \times 39.69$$ $$u \approx 3.51 \times 10^{-10} \text{ J/m}^3$$ Final Answer: (a) The magnetic field vector is $\vec{B} = 2.1 \times 10^{-8} \hat{k} \text{ T}$. (b) The energy density of the electromagnetic wave is $3.51 \times 10^{-10} \text{ J/m}^3$.

Question

A plane electromagnetic wave of frequency $25 	ext{ MHz}$ travels in free space along the x-direction. At a particular point in space and time, the electric field is $\vec{E} = 6.3 \hat{j} 	ext{ V/m}$. (a) Calculate the magnitude and direction of the magnetic field vector $\vec{B}$ at this point. (b) Calculate the energy density of the electromagnetic wave.

Accepted Answer

Given Data: Frequency of the electromagnetic wave ($
u$) = $25 	ext{ MHz} = 25 	imes 10^6 	ext{ Hz}$ Direction of propagation = x-direction ($i$) Electric field vector ($\vec{E}$) = $6.3 \hat{j} 	ext{ V/m}$ Permeability of free space ($\mu0$) = $4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}$ Permittivity of free space ($\varepsilon0$) = $8.854 	imes 10^{-12} 	ext{ C}^2/(	ext{N}\cdot	ext{m}^2)$ Speed of light ($c$) = $3 	imes 10^8 	ext{ m/s}$ Part (a): Magnitude and Direction of Magnetic Field Vector ($\vec{B}$) Magnitude Calculation: The relationship between the amplitude of the electric field ($E0$) and magnetic field ($B0$) is given by: $$B0 = \frac{E0}{c}$| Substituting the given values: $$B0 = \frac{6.3 	ext{ V/m}}{3 	imes 10^8 	ext{ m/s}} = 2.1 	imes 10^{-8} 	ext{ Tesla (T)}$$ Direction Calculation: The direction of propagation of an electromagnetic wave is given by the direction of the cross product vector $\vec{E} 	imes \vec{B}$. Since the wave travels along the x-axis ($+\hat{i}$) and the electric field is along the y-axis ($+\hat{j}$): $$\hat{i} = \hat{j} 	imes \hat{k}$| Therefore, the magnetic field $\vec{B}$ must point along the z-axis ($+\hat{k}$). Thus, $\vec{B} = 2.1 	imes 10^{-8} \hat{k} 	ext{ T}$. Part (b): Energy Density of the Electromagnetic Wave
The total energy density ($u$) of an electromagnetic wave is the sum of electric energy density ($uE$) and magnetic energy density ($uB$), which are equal to each other ($uE = uB$). Formula for Energy Density: $$u = \varepsilon0 E0^2$$ Calculation: $$u = (8.854 	imes 10^{-12}) 	imes (6.3)^2$$ $$u = 8.854 	imes 10^{-12} 	imes 39.69$$ $$u \approx 3.51 	imes 10^{-10} 	ext{ J/m}^3$$ Final Answer: (a) The magnetic field vector is $\vec{B} = 2.1 	imes 10^{-8} \hat{k} 	ext{ T}$. (b) The energy density of the electromagnetic wave is $3.51 	imes 10^{-10} 	ext{ J/m}^3$.

Given Data:

Part (a): Magnitude and Direction of Magnetic Field Vector ($\vec{B}$)

Part (b): Energy Density of the Electromagnetic Wave\nThe total energy density ($u$) of an electromagnetic wave is the sum of electric energy density ($u_E$) and magnetic energy density ($u_B$), which are equal to each other ($u_E = u_B$).

Final Answer: